Density= mass/volume
16.39g/18.00mL = 0.9105 g/mL
Make sure to use the correct number of significant figures (4)
Hello! Your answer is C, one atom of iron and three atoms of chlorine.
In this case, there is some ambiguity with "Cl". It could be C + I, or Cl. (One being carbon + iodine and the other Chlorine). However, we can see that the only choice that mentions iodine states two atoms, which is incorrect, as if it was iodine, the 3 indicates that there would be three atoms.
Therefore, Cl must be chlorine. If you look at choice C, there are three atoms of chlorine, as indicated by the 3 at the end of the Cl. There is also one molecule of Fe, as there is no number at the end of it so there has to be only one.
B is not correct, as carbon would be only "C". A is not correct as a molecule is a collection of atoms bonded together, therefore, this compound could be named a molecule, but there are not multiple molecules inside this compound.
Hope this helps!
Explanation:
P1V1 = nRT1
P2V2 = nRT2
Divide one by the other:
P1V1/P2V2 = nRT1/nRT2
From which:
P1V1/P2V2 = T1/T2
(Or P1V1 = P2V2 under isothermal conditions)
Inverting and isolating T2 (final temp)
(P2V2/P1V1)T1 = T2 (Temp in K).
Now P1/P2 = 1
V1/V2 = 1/2
T1 = 273 K, the initial temp.
Therefore, inserting these values into above:
2 x 273 K = T2 = 546 K, or 273 C.
Thus, increasing the temperature to 273 C from 0C doubles its volume, assuming ideal gas behaviour. This result could have been inferred from the fact that the the volume vs temperature line above the boiling temperature of the gas would theoretically have passed through the origin (0 K) which means that a doubling of temperature at any temperature above the bp of the gas, doubles the volume.
From the ideal gas equation:
V = nRT/P or at constant pressure:
V = kT where the constant k = nR/P. Therefore, theoretically, at 0 K the volume is zero. Of course, in practice that would not happen since a very small percentage of the volume would be taken up by the solidified gas.
The properties of the electrons of an atom are determined in large part by the number of protons presented in the nucleus of the atom
The average atomic mass of tellurium, calculated from its eight isotopes (Te-120 (0.09%), Te-122 (2.46%), Te-123 (0.87%), Te-124 (4.61%), Te-125 (6.99%), Te-126 (18.71%), Te-128 (31.79%), and Te-130 (34.48%)) is 127.723 amu.
The average atomic mass of Te can be calculated as follows:
Where:
m: is the mass
%: is the abundance percent
Knowing all the masses and abundance values, we have:
To find the <u>average atomic mass</u> we need to change all the <u>percent values</u> to <u>decimal ones</u>
Therefore, the average atomic mass of tellurium is 127.723 amu.
You can find more about average atomic mass here brainly.com/question/11096711?referrer=searchResults
I hope it helps you!
